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Unformatted text preview: Pstat160A Winter 2009 Solution Homework #2 Problem 1. Part 1) It is clear that X and Y are not independent, for instance P ( X = 1) = P ( Y = 1) = 1 4 but P ( X = 1 ,Y = 1) = 6 = 1 4 × 1 4 . By symmetry about 0, it’s clear that E ( X ) = E ( Y ) = 0. Therefore Cov( X,Y ) = E ( XY ). But for each point, the either X or Y is 0 (the points are on the axis). Therefore E ( XY ) = 0 ⇒ Cov( X,Y ) = 0 ⇒ ρ ( X,Y ) = 0. So X and Y are uncorrelated, but not independent. The issue here is that there is no “preferred line” where the points tends to be close, any line going through the origin is as good as any others. Part 2) a) By symmetry as in part 1) E ( X ) = E ( Y ) = 0. or more formally we can do as follows: the pdf of X is given by: p X ( 1) = p X,Y ( 1 , 1) = 1 / 5 , p X (0) = p X,Y (0 , 1) + p X,Y (0 , 0) + p X,Y (0 , 1) = 1 / 5 + 1 / 5 + 1 / 5 = 3 / 5 p X (1) = p X,Y (1 , 1) = 1 / 5 So indeed E ( X ) = 1 × 1 / 5 + 0 × 3 / 5 + 1 × 1 / 5 = 0. The pdf of Y is p Y ( 1) = 2 / 5 , p Y (0) = 1 / 5 , p Y (1) = 2 / 5 and we get E ( Y ) = 1 × 2 / 5 + 0 × 2 / 5 + 1 × 2 / 5 = 0 Therefore Cov(...
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 Spring '08
 EduardoMontoya
 Covariance, var, Covariance and correlation, Multivariate normal distribution, Cov

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